If $AT=TA$ with $A\geq0$. Why $A^{1/2}T=TA^{1/2}$?

Question

Let $\mathcal{H}$ be a complex Hilbert space. Let $T\in \mathcal{B}(\mathcal{H})$ and let $A\in \mathcal{B}(\mathcal{H})^+$ (i.e. $A^*=A$ and $\langle Ax\;| \;x\rangle \geq0,\;\forall x\in \mathcal{H}$).

Assume that $AT=TA$. Why $A^{1/2}T=TA^{1/2}$?

Thank you

Answer 1

You can check that $p(A)T=Tp(A)$ for any polynomial $p$.

For any continuous function $f$ on $\sigma(A)$, there is a sequence $\{p_n\}$ of polynomials such that $p_n$ is convergent to $f$ uniformly. Hence $p_n(A)$ is convergent to $f(A)$ and thus $f(A)T=Tf(A)$.

Specially, $f(x)=\sqrt{x}, x\in \sigma(A)\subset [0,\infty)$ is your case.

Answer 2

Looking at the construction of the square-root operator, the commutation property in question is a somewhat simple consequence. The following argument is taken from Section 104 of the book "Functional Analysis" by Riesz & Nagy (1990, Dover publications).

Let $A\in\mathcal B(\mathcal H)^+$ be given where $\mathcal H$ is any complex Hilbert space, we are looking for a solution of the equation $X^2=A$. We define

$$ B=\operatorname{id}_{\mathcal H}-\frac{A}{\Vert A\Vert}\in\mathcal B(\mathcal H)^+\qquad\text{ and }\qquad Y=\operatorname{id}_{\mathcal H}-\frac{X}{\sqrt{\Vert A\Vert}} $$

(if $A=0$, all of this is trivial). Now $X^2=A$ becomes

$$ Y=\frac12(B+Y^2).\tag*{$(1)$} $$

as can be seen easily. Defining the sequence $(Y_n)_{n\in\mathbb N_0}$ via $Y_0=0, Y_1=\frac12B$ and $Y_{n+1}=\frac12(B+Y_n^2)$, it can be shown that $(Y_n)_{n\in\mathbb N_0}$ converges to $Y$ in the strong sense where $Y$ is a solution to (1) so $Y=\sqrt{A}$.

By this construction, every $Y_n$ is a polynomial in $B$ (and thus in $A$) so if some operator $T\in\mathcal B(\mathcal H)$ commutes with $A$, it commutes with every $Y_n$ and thus with its strong limit $Y=\sqrt{A}$.